91Ó°ÊÓ

The first step in understanding second-order linear differential equations is to focus on the characteristic equation, which helps us find the nature of the solution. These types of differential equations typically have the form:

• \( \frac{d^2 x}{dt^2} + a \frac{dx}{dt} + bx = 0 \)

Here, we assume a solution of the form \( x(t) = e^{rt} \), where \( r \) is a constant we need to find. Substituting this into the differential equation forms a quadratic equation known as the characteristic equation, given by:

• \( r^2 + ar + b = 0 \)

In our original exercise, this becomes \( r^2 + 2br + k^2 = 0 \). Solving this equation is crucial as it helps determine whether the roots are real or complex, which directly impacts the form of the solution.

When the roots of the characteristic equation turn out to be complex, it introduces oscillatory behavior to the solution. A complex root occurs when the discriminant of the characteristic equation is negative. This results in roots of the form \( r = -b \pm i\omega \), where \( \omega = \sqrt{k^2 - b^2} \).In the context of our differential equation, since the condition \( k > b \) is satisfied, the discriminant \( b^2 - k^2 \) is negative, confirming that the roots are complex:

• \( r_1 = -b + i\omega \)
• \( r_2 = -b - i\omega \)

These complex roots lead to a general solution that involves trigonometric functions due to Euler's formula, resulting in a solution of the form:

• \( x(t) = e^{-bt}(C_1 \cos(\omega t) + C_2 \sin(\omega t)) \)

This highlights the damping and oscillatory nature within the system, frequently seen in physics and engineering contexts.

When solving differential equations, initial conditions are vital for finding a unique particular solution. They are generally provided as values of the function and its derivatives at a specific point, usually \( t = 0 \). These conditions help determine unknown coefficients in the general solution.In our exercise, the initial conditions are:

• \( x(0) = 0 \)
• \( \frac{dx}{dt}(0) = v_0 \)

The condition \( x(0) = 0 \) allows us to solve for \( C_1 \), giving \( C_1 = 0 \). The derivation of the solution's first derivative with respect to time and the second condition \( \frac{dx}{dt}(0) = v_0 \) helps us solve for \( C_2 \). Upon substitution, we derive that \( C_2 = \frac{v_0}{\omega} \).By incorporating these constants back into the general solution, the particular solution becomes:

• \( x(t) = \frac{v_0}{\omega} e^{-bt} \sin(\omega t) \)

This particular solution describes the specific response of the system, ensuring it fits the given initial circumstances.